[题目链接] http://acm.hdu.edu.cn/showproblem.php?pid=5755 [题目大意] 一个n*m由0,1,2组成的矩阵,每次操作可以选取一个方格,使得它加上2之后对3取模,周围的四个方格加上1后对3取模,请你在n*m操作次数内让整个矩阵变成0.输出一种方案. [题解] 枚举第一行的方式显然是不行的,因为3的30次方显然不是可以承受的范围,考虑如果存在第0行元素,那么这一行的最终状态就是第一行的操作次数,因为每个格子很明显只会由第一行对应的正下方的格子影响,我们…

http://acm.hdu.edu.cn/showproblem.php?pid=5755 题意: n*m矩阵,每个格有数字0/1/2 每选择一个格子,这个格子+2,4方向相邻格子+1 如何选择格子,可以使每个格子的数最后 %3=0 #include<cstdio> #include<cstring> #include<algorithm> using namespace std; ; int n,m,t; ][]; ]; int turn(int i,int j)…

题目链接 给n*m的方格, 每个格子有值{0, 1, 2}. 然后可以对格子进行操作, 如果选择了一个格子, 那么这个格子的值+2, 这个格子上下左右的格子+1, 并且模3. 问你将所有格子变成0的操作方法. 其实就是一个模3的方程组, 高斯消元就可以了. 不知道为什么昨天比赛就是想不到...... #include <iostream> #include <vector> #include <cstdio> #include <cstring> #incl…

可以设n*m个未知量,建立n*m个方程.位置i,j可以建立方程 (2*x[i*m+j]+x[(i-1)*m+j]+x[(i+1)*m+j]+x[i*m+j-1]+x[i*m+j+1])%3=3-b[i][j]; 套了个高斯消元的板子过了. #pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cmath>…

Gambler Bo Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1152    Accepted Submission(s): 471Special Judge Problem Description Gambler Bo is very proficient in a matrix game. You have a N×M m…

http://acm.hdu.edu.cn/showproblem.php?pid=5755 题意:一个N*M的矩阵,改变一个格子,本身+2,四周+1.同时mod 3;问操作多少次,矩阵变为全0.输出次数和具体位置 由于影响是相互的,所以增广矩阵的系数a[t][t+1] 或者是 a[t+1][t]均可:只需注意往结果中添加位置时,x[i]表示要操作x[i]次,所以位置要重复添加x[i]次: 高斯消元时间复杂度为O(N3M3); #pragma comment(linker, "/STACK:10…

pro:给定N*M的矩阵,每次操作一个位置,它会增加2,周围4个位置会增加1.给定初始状态,求一种方案,使得最后的数都为0:(%3意义下. sol:(N*M)^3的复杂度的居然过了.          好像标程是M^3的,因为第一排确定了,后面的都确定了.所以我们只要设关于第一排的方程,那么跑下来,第N+1排的都为0,则合法. (此题由于3的特殊性,x关于3的逆元=x:所以不用求逆元 #include<bits/stdc++.h> #define rep(i,a,b) for(int i=a;…

Sqrt Bo 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5752 Description Let's define the function f(n)=⌊n−−√⌋. Bo wanted to know the minimum number y which satisfies fy(n)=1. note:f1(n)=f(n),fy(n)=f(fy−1(n)) It is a pity that Bo can only use 1 unit…

Permutation Bo 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5753 Description There are two sequences h1∼hn and c1∼cn. h1∼hn is a permutation of 1∼n. particularly, h0=hn+1=0. We define the expression [condition] is 1 when condition is True,is 0 whe…

Teacher Bo 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5762 Description Teacher BoBo is a geography teacher in the school.One day in his class,he marked N points in the map,the i-th point is at (Xi,Yi).He wonders,whether there is a tetrad (A,B,C,…

Sqrt Bo Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 2221    Accepted Submission(s): 882 Problem Description Let's define the function f(n)=⌊n−−√⌋. Bo wanted to know the minimum number y wh…

Rower Bo 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5761 Description There is a river on the Cartesian coordinate system,the river is flowing along the x-axis direction. Rower Bo is placed at (0,a) at first.He wants to get to origin (0,0) by boa…

Teacher Bo 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5762 Description Teacher BoBo is a geography teacher in the school.One day in his class,he marked N points in the map,the i-th point is at (Xi,Yi).He wonders,whether there is a tetrad (A,B,C,…

Permutation Bo 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5753 Description There are two sequences h1∼hn and c1∼cn. h1∼hn is a permutation of 1∼n. particularly, h0=hn+1=0. We define the expression [condition] is 1 when condition is True,is 0 whe…

Sqrt Bo 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5752 Description Let's define the function f(n)=⌊n−−√⌋. Bo wanted to know the minimum number y which satisfies fy(n)=1. note:f1(n)=f(n),fy(n)=f(fy−1(n)) It is a pity that Bo can only use 1 unit…

传送门 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Special Judge Problem DescriptionThere is a river on the Cartesian coordinate system,the river is flowing along the x-axis direction. Rower Bo is placed at $(0,a…

Teacher Bo Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1014    Accepted Submission(s): 561 Problem Description Teacher BoBo is a geography teacher in the school.One day in his class,he mar…

1010 Rower Bo 首先这个题微分方程强解显然是可以的,但是可以发现如果设参比较巧妙就能得到很方便的做法. 先分解v_1v​1​​, 设船到原点的距离是rr,容易列出方程 \frac{ dr}{ dt}=v_2\cos \theta-v_1​dt​​dr​​=v​2​​cosθ−v​1​​ \frac{ dx}{ dt}=v_2-v_1\cos \theta​dt​​dx​​=v​2​​−v​1​​cosθ 上下界都是清晰的,定积分一下: 0-a=v_2\int_0^T\cos\thet…

这里是一个比较简单的问题:考虑每个数对和的贡献.先考虑数列两端的值,两端的摆放的值总计有2种,比如左端:0,大,小:0,小,大:有1/2的贡献度.右端同理. 中间的书总计有6种可能.小,中,大.其中有两种对答案有贡献,即1/3的贡献度.加和计算可得到答案. Permutation Bo Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s):…

Sqrt Bo Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 980    Accepted Submission(s): 452 Problem Description Let's define the function f(n)=⌊n√⌋. Bo wanted to know the minimum number y which…

Teacher Bo Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 752    Accepted Submission(s): 412 Problem Description Teacher BoBo is a geography teacher in the school.One day in his class,he mark…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5761 题目大意: 船在(0,a),船速v1,水速v2沿x轴正向,船头始终指向(0,0),问到达(0,0)用时,无解输出Infinity. 题目思路: [数学] 说是数学其实更像物理. 很明显v1<=v2时无解. 这道题列积分方程然后解出来式子是t=a*v1/(v12-v22). 我列积分方程的时候少了一个式子没解出来. 可以看看半根毛线的.http://images2015.cnblogs.com…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5762 题目大意: 给n个点,坐标范围0~m(n,m<=105),求是否存在2个点对满足哈夫曼距离相等. 题目思路: [模拟] 乍一看n2绝对T了,但是细想之下发现,坐标范围只有105,那么哈夫曼距离最多就2x105种,所以当循环超出这个范围时肯定能找到解(抽屉原理). // //by coolxxx // #include<iostream> #include<algorithm&g…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5752 题目大意: 定义f(n)=⌊√n⌋,fy(n)=f(fy-1(n)),求y使得fy(n)=1.如果y>5输出TAT.(n<10100) 题目思路: [模拟] 5层迭代是232,所以特判一下层数是5的,其余开根号做.注意数据有0. 队友写的. #include<stdio.h> #include<string.h> #include<math.h> int…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5753 题目大意: 两个序列h和c,h为1~n的乱序.h[0]=h[n+1]=0,[A]表示A为真则为1,假为0. 函数f(h)=(i=1~n)∑ci[hi>hi−1 && hi>hi+1] 现在给定c的值,求f(h)的期望. 题目思路: [数学] 头尾的概率为1/2,中间的概率为1/3,直接求和. // //by coolxxx // #include<iostream&g…

[题目链接] http://acm.hdu.edu.cn/showproblem.php?pid=5758 [题目大意] 给出一棵树,每条路长度为1,允许从一个节点传送到任意一个节点,现在要求在传送次数尽量少的情况下至少经过每条路一遍啊,同时最小化走过的路程总长度.输出路程总长度. [题解] 首先,对于传送次数尽量少这个条件,我们很容易发现,当且仅当每次出发点和终止点都是叶节点的时候,是最少的,当然在叶节点无法两两匹配的时候,再多走一条链. 然后就是叶节点的匹配问题,使得匹配后的叶节点连线覆盖全…

Swipe Bo Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 455    Accepted Submission(s): 111 Problem Description “Swipe Bo” is a puzzle game that requires foresight and skill. The main character…

Life Winner Bo Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1019    Accepted Submission(s): 373 Problem Description Bo is a "Life Winner".He likes playing chessboard games with his gir…

Rower Bo Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 641    Accepted Submission(s): 199Special Judge Problem Description ,and the speed of the water flow is v2.He will adjust the direction…

题目:传送门. 题意:平面上有n个点,问是否存在四个点 (A,B,C,D)(A<B,C<D,A≠CorB≠D)使得AB的横纵坐标差的绝对值的和等于CD的横纵坐标差的绝对值的和,n<10^5,点的坐标值m<10^5. 题解:表面上这道题复杂度是O(n^2)会超时的,而实际上这些坐标差绝对值的和最大是2*10^5,所以复杂度不是O(n^2),而是O(min(n^2,m)),这就是著名的鸽笼原理. #include <iostream> #include <cstdio…